hdu 1385 Minimum Transport Cost

点击打开链接hdu 1385

思路：最短路+SPFA+路径记录+路径字典序比较

分析：
1 题目要求的单源的最短路，所以可以选择任意一种单源最短路来求解
2 题目还要求在路径和相同情况下要字典序小的，那么就要在更新dis数组的时候进行更新路径，如果是dis[i]>tmp,那么直接更新；如果是dis[i] == tmp的情况下，那么就要求出star->i 和 star->x的路径进行比较，然后判断能否更新.

3 注意询问的时候可能问的是同一个点。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

#define MAXN 110
#define INF 0xFFFFFFF

int n , star , end , pos;
int value[MAXN][MAXN];
int tmp_value[MAXN][MAXN];
int dis[MAXN];
int charge[MAXN];
int vis[MAXN];
int route[MAXN];
int father[MAXN];
int tmp_charge[MAXN];
queue<int>q;

/*初始化数组*/
void init(){
   for(int i = 1 ; i <= n ; i++){
      for(int j = 1 ; j <= n ; j++)
         value[i][j] = INF;
      value[i][i] = 0;
   }
}

/*dfs找路径*/
void dfs(int num , char *str){
     if(num == -1)
       return;
     dfs(father[num] , str);
     str[pos++] = num+'0';
}

void SPFA(int s){
   memset(vis , 0 , sizeof(vis));
   for(int i = 1 ; i <= n ; i++){
      dis[i] = INF;
      father[i] = -1;
   }
   vis[s] = 1;
   dis[s] = 0;
   q.push(s);
   while(!q.empty()){
       int x = q.front();
       q.pop();
       vis[x] = 0;
       for(int i = 1 ; i <= n ; i++){
          int tmp = dis[x] + tmp_value[x][i] + tmp_charge[x];
          /*dis[i]>tmp直接更新*/
          if(dis[i] > tmp){
             dis[i] = tmp;
             father[i] = x;
             if(!vis[i]){
                vis[i] = 1;
                q.push(i);
             }
          }
          /*dis[i] == tmp的时候求出路径*/
          else if(dis[i] == tmp && x != i && dis[i] != INF){        
             char ch1[MAXN] = {'\0'} , ch2[MAXN] = {'\0'};
             pos = 0;
             dfs(i , ch1);
             pos = 0;
             dfs(x , ch2);
             ch2[pos] = x+'0';/*这个地方注意就是x要加上去*/
             /*如果字典序小就要更新*/
             if(strcmp(ch1 , ch2) > 0){
                father[i] = x;
                if(!vis[i]){
                  vis[i] = 1;
                  q.push(i);
                }
             }
          }
       }
   }
}

int main(){
   int x , y , num , cnt , ans;
   while(scanf("%d" , &n) && n){
      init();
      /*输入边*/
      for(int i = 1 ; i <= n ; i++){
         for(int j = 1 ; j <= n ; j++){
            scanf("%d" , &num);  
            if(num != -1 && value[i][j] > num)
              value[i][j] = num;
         }
      }
      /*输入过路费*/
      for(int i = 1 ; i <= n ; i++)
         scanf("%d" , &charge[i]);
      /*询问*/
      while(1){
         scanf("%d%d" , &star , &end);
         if(star == -1 && end == -1)
           break;
         /*数组复制*/
         memcpy(tmp_charge , charge , sizeof(charge));
         memcpy(tmp_value , value , sizeof(value));
         tmp_charge[star] = tmp_charge[end] = 0;/*起点和终点的过路费为0*/
         
         SPFA(star);
         
         /*输出*/
         printf("From %d to %d :\n" , star , end);
         printf("Path: ");
         
         memset(route , 0 , sizeof(route));
         ans = dis[end];

         if(star != end){
            cnt = 0;
            x = end;
            while(1){
               y = father[x];
               route[cnt++] = y;
               if(y == star)
                 break;
               x = y;
            }
            for(int i = cnt-1 ; i >= 0 ; i--)
              printf("%d-->" , route[i]);
         }

         printf("%d\n" , end);
         printf("Total cost : %d\n\n" , ans);
      }
   }
   return 0;
}

2 floyd  ，由于floyd是每一次都用k去更新dis，所以我们只要在更新的同时更新path即可，由于k是从1-n的，那么这样最后的path肯定是字典序最小的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define MAXN 110
#define INF 0xFFFFFFF

int n;
int value[MAXN][MAXN];
int charge[MAXN];
int path[MAXN][MAXN];

void init(){
   for(int i = 1 ; i <= n ; i++){
      for(int j = 1 ; j <= n ; j++){
         value[i][j] = INF;
         path[i][j] = j;
      }
      value[i][i] = 0;
   }
}

void floyd(){
    for(int k = 1 ; k <= n ; k++){
       for(int i = 1 ; i <= n ; i++){
          for(int j = 1 ; j <= n ; j++){
            int tmp = value[i][k]+value[k][j]+charge[k];
            if(value[i][j] > tmp){
              value[i][j] = tmp;
              path[i][j] = path[i][k];
            }
            else if(value[i][j] == tmp && path[i][j] > path[i][k])
              path[i][j] = path[i][k];
          }
       }
    }
}

int main(){
   int x , y , num , star , end;
   while(scanf("%d" , &n) && n){
      init();
      for(int i = 1 ; i <= n ; i++){
         for(int j = 1 ; j <= n ; j++){
            scanf("%d" , &num);
            if(num != -1)
              value[i][j] = num;
         }
      }
      for(int i = 1 ; i <= n ; i++)
         scanf("%d" , &charge[i]);

      floyd();

      while(1){
         scanf("%d%d" , &star , &end);
         if(star == -1 && end == -1)
           break;

         printf("From %d to %d :\n" , star , end);
         printf("Path: %d" , star);
         x = star , y = end;
         if(star != end){
            while(1){
                int p = path[x][y];
                printf("-->%d" , p);
                if(p == end)
                   break;
                x = p;
            }
         }
         printf("\n");
         printf("Total cost : %d\n\n" , value[star][end]);
      }
   }
   return 0;
}